# Heron's Formula

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Heron's Formula (sometimes called Hero's formula) is a formula for finding the area of a triangle given only the three side lengths.

## Theorem

For any triangle with side lengths ${a}, {b}, {c}$, the area ${A}$ can be found using the following formula: $A=\sqrt{s(s-a)(s-b)(s-c)}$

where the semi-perimeter $s=\frac{a+b+c}{2}$.

## Proof $[ABC]=\frac{ab}{2}\sin C$ $=\frac{ab}{2}\sqrt{1-\cos^2 C}$ $=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}$ $=\sqrt{\frac{a^2b^2}{4}\left[1-\frac{(a^2+b^2-c^2)^2}{4a^2b^2}\right]}$ $=\sqrt{\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}$ $=\sqrt{\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16}}$ $=\sqrt{\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}$ $=\sqrt{\frac{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}{16}}$ $=\sqrt{s(s-a)(s-b)(s-c)}$

## Isosceles Triangle Simplification $A=\sqrt{s(s-a)(s-b)(s-c)}$ for all triangles $b=c$ for all isosceles triangles $A=\sqrt{s(s-a)(s-b)(s-b)}$ simplifies to $A=(s-b)\sqrt{s(s-a)}$ $\blacksquare$

## Example

Let's say that you have a right triangle with the sides $3$ , $4$ , and $5$. Your semi- perimeter would be $6$ since $(3+4+5)$ ÷ $2$ is $6$. Then you have $6-3=3$, $6-4=2$, $6-5=1$. $1+2+3=6.$ $6\cdot 6 = 36$ The square root of $36$ is $6$. The area of your triangle is $6$.